PHY 317K
Homework 6 Solutions
1.
B
For a circle (like a coin),
A
=
πr
2
=
πd
2
/
4, where
d
is the diameter.
If
d
changes
by Δ
d
, the area becomes
A
+ Δ
A
=
π
(
d
+ Δ
d
)
2
/
4 =
πd
2
/
4 + 2
πd
Δ
d/
4 +
π
(Δ
d
)
2
/
4.
Therefore, Δ
A
= 2
πd
Δ
d/
4, if we ignore the third term since Δ
d
is very small compared
to
d
. Then if we multiply the left side by
d/d
, we see Δ
A
= 2(
πd
2
/
4)Δ
d/d
, from which
it follows: Δ
A/A
= 2Δ
d/d
. Hence, a 0.17% change in
d
results in a 0.34% change in
the area of the coin.
2.
C
For coefficient of linear expansion
α
,
Δ
V
V
= 3
α
Δ
T
. Thus, here
Δ
V
= 3
αV
Δ
T
= 3(1
.
1
×
10

5
/C
◦
)(4
.
00 cm)
3
(57 C
◦
) = 0
.
12 cm
3
.
3.
D
The heat capacity at constant volume and the heat capacity at constant pressure have
different values because the system does work at constant pressure, but not at constant
volume. I.e., when the volume changes, work is done.
4.
A
The energy needed to melt
A
is
ML
, where
M
is the mass of
A
and
L
is latent heat of
fusion. The total heat energy that
A
receives is
C
A
(
T
final

T
A
) +
ML
, and this must
equal the heat energy lost by
B
:
C
B
(
T
B

T
final
) =
C
A
(
T
final

T
A
) +
ML.
Solving for
T
final
we find:
T
final
=
C
A
T
A
+
C
B
T
B

ML
C
A
+
C
B
5.
A
During the time that latent heat is involved in a change of phase the temperature does
not change.
6.
E
Several steps are involved:
1) The ice is raised from

20
◦
C to 0
◦
C:
E
1
= 10 g(0
.
5 cal
/
g
·
C
◦
)(20 C
◦
) = 100 cal.
2) The ice is melted:
E
2
= 10 g(80 cal
/
g) = 800 cal.
3) The water is heated to 100
◦
C:
E
3
= 10 g(1
.
00 cal
/
g
·
C
◦
)(100 C
◦
) = 1000 cal.
4) The water is vaporized:
E
4
= 10 g(540 cal
/
g) = 5400 cal.
5) The steam is heated to 130
◦
C:
E
5
= 10 g(0
.
5 cal
/
g
·
C
◦
)(60 C
◦
) = 300 cal.
The sum of the energies required for all steps is 7600 cal.
1
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7.
C
The humidifier heats 15 g of water from 20
◦
C to 100
◦
C and then evaporates those 15 g
each minute. The energy needed for that is:
E
=
mc
Δ
T
+
mL
V
= (15 g)
·
(4190 J
/
kg K)
·
(80 K) + (15 g)
·
(2256 kJ
/
kg) = 39000 J
where
L
V
is heat of vaporization.
8.
B
According to the first law of thermodynamics, applied to gas, the increase in the
internal energy during the process is equal to the heat input plus the work done
on
the gas.
9.
D
The work done by the gas is equal to
W
=
pdV
.
Because of the direction of the
processes, we see that I and II result in negative work, while III, IV, and V result in
positive work. In magnitude, the positive work in IV is largest. This is simply based
on the area under the curves.
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 Spring '07
 KOPP
 Thermodynamics, Energy, Entropy

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